# In Fig. 6.14, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.

It is given to us –

DE || QR and n is the transversal intersecting DE and QR at points A and B respectively.

We have to find APB.

AP and BP are the bisectors of EAB and ABR respectively.

EAP = PAB

EAB = 2 × PAB - - - - (i)

Also, RBP = PBA

RBA = 2 × PBA - - - - (ii)

Since, DE || QR,

RBA = EAn (Corresponding angles)

RBA = 180° - EAB (By linear pair axiom, EAn + EAB = 180°)

Using equations (i) and (ii) in the above equation,

2 × PBA = 180° - (2 × PAB)

PBA = 90° - PAB (Dividing both sides by 2)

PBA + PAB = 90° - - - - (iii)

Now, in ΔPAB,

PBA + PAB + APB = 180° (Sum of three angles of a triangle is 180°)

90° + APB = 180° [From equation (iii)]

APB = 90°

Thus, the value of APB is 90°.

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